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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
<li><a href="sec2_1.html" data-scroll="sec2_1" class="internal">Linear Equations</a></li>
<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
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<li><a href="sec3_2.html" data-scroll="sec3_2" class="active">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
<li><a href="sec3_4.html" data-scroll="sec3_4" class="internal">Complex roots of the characteristic equations</a></li>
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<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
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<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec3_2"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">3.2</span> <span class="title">Fundamental Solutions of Linear Homogeneous Equations</span>
</h2>
<p id="p-70">Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime \prime}+p(x) y^{\prime}+q(x) y=g(x),\quad y(x_0)=y_0,~y^{\prime}(x_0)=y_1.
\end{equation*}
</div>
<p id="p-71"><dfn class="terminology">Theorem</dfn> If <span class="process-math">\(p(x)\text{,}\)</span> <span class="process-math">\(q(x)\)</span> and <span class="process-math">\(g(x)\)</span> are continuous in an open interval <span class="process-math">\(I\text{:}\)</span> <span class="process-math">\(\alpha&lt;x&lt;\beta\text{,}\)</span> which contains <span class="process-math">\(x_0\text{,}\)</span> then the initial value problem has a unique solution which is valid in <span class="process-math">\(I\text{.}\)</span></p>
<p id="p-72"><dfn class="terminology">Example 1:</dfn> Determine the largest intervals in which the following initial value problem has a unique solution.</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
x (x-4) y^{\prime \prime}+3 x y^{\prime}+4 y=2,\quad y(3)=0,\quad y^{\prime}(3)=1.
\end{equation*}
</div>
<p id="p-73"><dfn class="terminology">Solution:</dfn> The equation can be arranged as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime \prime}+\frac{3 x}{x (x-4)} y^{\prime}+\frac{4}{x (x-4)} y=\frac{2}{x(x-4)}.
\end{equation*}
</div>
<p class="continuation"><span class="process-math">\(p(x)\text{,}\)</span> <span class="process-math">\(q(x)\)</span> and <span class="process-math">\(g(x)\)</span> are continuous for <span class="process-math">\(x \neq 0\)</span> and <span class="process-math">\(x \neq 4\text{,}\)</span> i.e., in <span class="process-math">\(-\infty&lt;x&lt;0\)</span> or <span class="process-math">\(0&lt;x&lt;4\)</span> or <span class="process-math">\(4&lt;x&lt;\infty\text{.}\)</span> The interval <span class="process-math">\(0&lt;x&lt;4\)</span> contains <span class="process-math">\(x=3\text{.}\)</span> Thus, it is the largest interval in which there is a unique solution.</p>
<p id="p-74"><dfn class="terminology">Theorem</dfn> If <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are two solutions to the differential equations <span class="process-math">\(y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0\text{,}\)</span> then the linear combination <span class="process-math">\(C_1 y_1+C_2 y_2\)</span> is also a solution for any values of <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\text{.}\)</span></p>
<p id="p-75"><dfn class="terminology">Proof</dfn></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;y_1~ \textrm{is a solution}:\quad y_1^{\prime \prime}+p(x) y_1^{\prime}+q(x) y_1=0,\\
&amp;y_2~ \textrm{is a solution}:\quad y_2^{\prime \prime}+p(x) y_2^{\prime}+q(x) y_2=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Let <span class="process-math">\(y=C_1 y_1+C_2 y_2\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
LHS&amp;=[C_1 y_1+C_2 y_2]^{\prime \prime}+p(x) [C_1 y_1+C_2 y_2]^{\prime}+q(x) [C_1 y_1+C_2 y_2]\\
&amp;=C_1 [y_1^{\prime \prime}+p(x) y_1^{\prime}+q(x) y_1]+C_2 [y_2^{\prime \prime}+p(x) y_2^{\prime}+q(x) y_2]\\
&amp;=0=RHS.
\end{aligned}
\end{equation*}
</div>
<p id="p-76"><dfn class="terminology">Question</dfn>: Whether <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> can be uniquely chosen to satisfy any initial conditions, say <span class="process-math">\(y(x_0)=a_0, ~y^{\prime}(x_0)=a_1\text{?}\)</span></p>
<p id="p-77"><dfn class="terminology">Answer</dfn>:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;y(x_0)=a_0:\quad a_0=C_1 y_1(x_0)+C_2 y_2 (x_0),\\
&amp;y^{\prime}(x_0)=a_1:\quad a_1=C_1 y_1^{\prime} (x_0)+C_2 y_2^{\prime} (x_0),\\
&amp;\rightarrow C_1=\frac{\left| \begin{array}{cc} a_0 &amp; y_2(x_0)\\ a_1  &amp; y_2^{\prime}(x_0) \end{array}    \right|}{W_0},\quad C_2=\frac{\left| \begin{array}{cc} y_1(x_0) &amp; a_0 \\ y_1^{\prime}(x_0) &amp;a_1 \end{array}    \right|}{W_0}, \quad (\mathrm{Cramer's~rule})
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
W_0=\left| \begin{array}{cc} y_1(x_0) &amp; y_2(x_0) \\ y_1^{\prime}(x_0) &amp; y_2^{\prime}(x_0) \end{array}    \right|.
\end{equation*}
</div>
<p id="p-78"><dfn class="terminology">Terminology</dfn>:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
W(y_1, y_2)=\left| \begin{array}{cc} y_1(x) &amp; y_2(x) \\ y_1^{\prime}(x) &amp; y_2^{\prime}(x) \end{array}    \right|
\end{equation*}
</div>
<p class="continuation">is called Wronskian for <span class="process-math">\(y_1(x)\)</span> and <span class="process-math">\(y_2(x)\text{.}\)</span></p>
<p id="p-79">One can observe that <span class="process-math">\(W_0=W|_{x=x_0}\text{.}\)</span> Thus we conclude that if <span class="process-math">\(W_0=W|_{x=x_0} \neq 0\text{,}\)</span> then <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> can be uniquely chosen to satisfy the initial conditions</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(x_0)=a_0,\quad y^{\prime}(x_0)=a_1.
\end{equation*}
</div>
<p id="p-80"><dfn class="terminology">Theorem</dfn> Suppose that <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are two solutions of</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_5_4.html">
\begin{equation*}
y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0.
\end{equation*}
</div>
<p class="continuation">If there is a point <span class="process-math">\(x_0\)</span> such that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_5_4.html" id="eq3_5_4">
\begin{equation}
W(y_1, y_2)|_{x=x_0}\neq 0\tag{3.2.1}
\end{equation}
</div>
<p class="continuation">then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_5_4.html">
\begin{equation*}
y=C_1 y_1(x)+C_2 y_2(x)
\end{equation*}
</div>
<p class="continuation">is the general solution. If (<a href="" class="xref" data-knowl="./knowl/eq3_5_4.html" title="Equation 3.2.1">(3.2.1)</a>) is satisfied, we call <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> form <dfn class="terminology">a fundamental set of solutions</dfn>.</p>
<p id="p-81"><dfn class="terminology">Example 1</dfn> Suppose that the characteristic equation of</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq3_6">
\begin{equation}
y^{\prime \prime}+b y^{\prime}+c y=0\tag{3.2.2}
\end{equation}
</div>
<p class="continuation">has two distinct real roots, <span class="process-math">\(r_1, r_2\text{.}\)</span> Show that <span class="process-math">\(y_1=e^{r_1 x}\text{,}\)</span> <span class="process-math">\(y_2=e^{r_2 x}\)</span> are a fundamental set of solutions.</p>
<p id="p-82"><dfn class="terminology">Proof:</dfn> We already know that <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are two solutions.</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq3_5_5">
\begin{equation}
W=\left| \begin{array}{cc} y_1 &amp; y_2 \\ y_1^{\prime} &amp; y_2^{\prime}     \end{array} \right|=
\left| \begin{array}{cc} e^{r_1 x} &amp; e^{r_2 x} \\ r_1 e^{r_1 x} &amp;r_2 e^{r_2 x}     \end{array} \right|
=r_2 e^{(r_1+r_2)x}-r_1 e^{(r_1+r_2)x} =(r_2-r_1) e^{(r_1+r_2)x} \neq 0.\tag{3.2.3}
\end{equation}
</div>
<p class="continuation">The above inequality holds for any <span class="process-math">\(x\text{.}\)</span> Thus <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are a fundamental set of solutions.</p>
<p id="p-83"><dfn class="terminology">Question:</dfn> What is the general solution?</p>
<p id="p-84"><dfn class="terminology">The general solution:</dfn> <span class="process-math">\(y=C_1 y_1+C_2 y_2=C_1 e^{r_1 x}+C_2 e^{r_2 x}\text{.}\)</span></p>
<p id="p-85"><dfn class="terminology">Proof:</dfn> Suppose that <span class="process-math">\(y=\phi(x)\)</span> is any solution of (<a href="" class="xref" data-knowl="./knowl/eq3_6.html" title="Equation 3.2.2">(3.2.2)</a>), we need to show that <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> can be found such that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
\phi(x)=C_1 y_1(x)+C_2 y_2(x).
\end{equation*}
</div>
<p class="continuation">Denote <span class="process-math">\(\phi(x_0)=a_0\text{,}\)</span> <span class="process-math">\(\phi^{\prime}(x_0)=a_1\text{.}\)</span> Then, the initial value problem</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html" id="eq3_7">
\begin{equation}
y^{\prime \prime}+p(x) y^{\prime}+q(x)y=0,\quad y(x_0)=a_0, \quad y^{\prime}(x_0)=a_1\tag{3.2.4}
\end{equation}
</div>
<p class="continuation">has a unique solution according to the theorem on the uniqueness and existence. Then <span class="process-math">\(y=\phi(x)\)</span> should be this unique solution (it satisfies both the ODE and initial conditions). On the other hand, we know that for the function</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
\bar{\phi}(x)=C_1 y_1(x)+C_2 y_2(x)
\end{equation*}
</div>
<p class="continuation">to satisfy (<a href="" class="xref" data-knowl="./knowl/eq3_7.html" title="Equation 3.2.4">(3.2.4)</a>), we only have to choose</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
C_1=\frac{\left| \begin{array}{cc} a_0 &amp; y_2(x_0)\\ a_1  &amp; y_2^{\prime}(x_0) \end{array}    \right|}{W_0},\quad C_2=\frac{\left| \begin{array}{cc} y_1(x_0) &amp; a_0 \\ y_1^{\prime}(x_0) &amp;a_1 \end{array}    \right|}{W_0},
\end{equation*}
</div>
<p class="continuation">where</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html" id="eq3_8">
\begin{equation}
W_0=\left| \begin{array}{cc} y_1(x_0) &amp; y_2(x_0) \\ y_1^{\prime}(x_0) &amp; y_2^{\prime}(x_0) \end{array}    \right|.\tag{3.2.5}
\end{equation}
</div>
<p class="continuation">Since <span class="process-math">\(y_1(x)\)</span> and <span class="process-math">\(y_2(x)\)</span> are a fundamental set of solutions as shown by (<a href="" class="xref" data-knowl="./knowl/eq3_5_5.html" title="Equation 3.2.3">(3.2.3)</a>), <span class="process-math">\(W_0\)</span> in (<a href="" class="xref" data-knowl="./knowl/eq3_8.html" title="Equation 3.2.5">(3.2.5)</a>) is nonzero. So <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> are uniquely determined. Therefore <span class="process-math">\(\bar{\phi}(x)\)</span> is also a solution to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq3_7.html" title="Equation 3.2.4">(3.2.4)</a>). We know both <span class="process-math">\(\bar{\phi}(x)\)</span> and <span class="process-math">\(\phi(x)\)</span> are solutions to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq3_7.html" title="Equation 3.2.4">(3.2.4)</a>) while this initial value problem has a unique solution. Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
\phi(x)=\bar{\phi}(x)=C_1 y_1(x)+C_2 y_2(x).
\end{equation*}
</div>
<p class="continuation">So <span class="process-math">\(y(x)=C_1 y_1(x)+C_2 y_2(x)\)</span> is the general solution.</p></section></div></main>
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